How much work did the movers do (horizontally) pushing a 130-kgkg crate 10.5 mm across a rough floor without acceleration, if the effective coefficient of friction was 0.50?

6.6885Joules

Explanation:

Wordone is expressed as the product of the moving force acting on the body and the distance.

Workdone = Force * distance

According to Newton second law;

Fm - Ff = ma

Since acceleration is zero;

Fm - Ff = 0

Fm  = Ff

Fm is the moving force

Ff is the frictional force

Since Ff = ηR

Fm = Ff = ηR

η is the coefficient of friction

R is the reaction

Get the moving force:

Fm =  ηR

Fm = ηmg

Fm = 0.50(130)(9.8)

Fm  = 637N

Recall that distance = 10.5mm = 0.0105m

Get the work done;

Work done = 637 * 0.0105

Work done = 6.6885Joules

Hence the movers did 6.6885Joules of work.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

what type of charge does it have? negative